Terms related to polygons - class-VIII

The least distance between the start and the end point can be calculated by the square sum root of the two distances travelled since north and west directions are perpendicular to each. hence, 
Least distance = $\sqrt{(15^2 + 8^2)}$
Least distance = 17 m

In simple closed curves, the shapes are closed by line-segments or by a curved line.

Triangle, quadrilateral, circle, etc., are examples of closed curves.

In order to draw a circle, first we need a compass and we need to take the required radius length with the help of a scale.

When we join no. of points without lifting our pen we make a shape which not necessarily should be straight or curve. We get a plane curve.
Therefore, A is the correct answer.

When the curve crosses itself it has end points. While a simple closed curve does not have any end points.
Therefore, B is the correct answer.

An open curve is a curve whose end points do not join. In other words it has 2 end points.
Therefore, C is the correct answer.

As the line segment has end points, it is an open curve.
Therefore, C is the correct answer.

The beginning point and the ending point in a closed curve is the same. There are no end points and starting points. Eg. Circle.
Therefore, B and C is the correct answer.

The statement is true.

A circle drawn on a plane divides into 3 parts:

Let $ x=\overline{BD} $ and let $ r$ be the radius of the small circle. 

For a circle x^2 + y^2 - 10x - 4y + 4 = 0, the center is (5, 2). Chords subtending a right angle at a point P(8, 2) pass through a fixed point. By the property of circles, if a chord subtends a right angle at P, the locus of the intersection of tangents at the chord's endpoints is related to the circle's geometry. The fixed point is the center (5, 2).

$m$ be the slope of tangent.

Given equation of curve is

${{e}^{2y}}=1+{{x}^{2}}$

Taking log both side and we get,

$ \log {{e}^{2y}}=\log \left( 1+{{x}^{2}} \right) $

$ 2y=\log \left( 1+{{x}^{2}} \right) $

On differentiating and we get,

$ 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}2x $

$ \Rightarrow \dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}} $

$m=\dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}}$

On put $x=\left( -1,1 \right)$

So,

$ m=\dfrac{1}{1+1}=\dfrac{1}{2} $

$ m=\dfrac{-1}{1+1}=\dfrac{-1}{2} $

Hence, the value of $m$ is $\left[-\dfrac{1}{2},\dfrac{1}{2} \right]$

Hence, this is the
answer.

A close curve is made up of a closed boundary. It initialised by a fixed point and end with the same point.

A curve that does not cross itself and ends at the same point where it begins.
Therefore, D is the correct answer.

$y^3=27x$

For the curves to touch, they must share a common tangent at a point. Setting y = 2(x-a)^2 and y = e^(2x) equal and their derivatives equal: 2(x-a)^2 = e^(2x) and 4(x-a) = 2e^(2x). Solving these leads to the condition for 'a'.